Re: [Phys-l] nifty question (solution)

[curtis osterhoudt <flutzpah@yahoo.com>]

The reason I'd asked "what happens if you pulled up, instead of down?", is that, 
for at least some range of up-pulls, I suspect the situation becomes unstable in 
equilibrium, and the cable will slip to one side or the other given something to 
break the left-right symmetry. I wondered if anyone might venture to show that, 
or disagrees with my guess. 


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________________________________
From: Carl Mungan <mungan@usna.edu>
To: phys-l@carnot.physics.buffalo.edu
Sent: Fri, September 24, 2010 6:00:35 AM
Subject: Re: [Phys-l] nifty question (solution)

Okay, looks like most people got it, and several people even 
anticipated the second question below. Reference for these questions 
is given in the last line below.

> Consider an ideal (ie. cannot stretch, and can bend freely) cable 
> suspended from two points of equal height (that are closer together 
> than the cable's length). The cable adopts the shape of a catenary 
> (hyperbolic cosine).
>
> Now suppose you hang a weight from the center of the cable. Does the 
> cable's center of mass move up, down, or remain at the same height?

The center of mass goes up. Here's a simple explanation:

Suppose instead of suddenly attaching the hanging mass m to the 
cable, I instead apply a downward force F to the center of the cable 
that slowly increases from 0 up to mg. (This guarantees the process 
is quasistatic, just like what we do in thermo when we compress a 
piston in a gas-filled cylinder.) Clearly I do positive work on the 
cable. Thus the energy of the cable increases. Since the process was 
quasistatic, there is never any KE of the cable. Further, the cable 
is ideal and thus doesn't stretch, so it cannot gain elastic PE. All 
that's left is for the cable to gain gravitational PE, and thus its 
COM must rise.

> From the same readings, here's another (much easier) nifty question. 
Suppose instead of pulling downward on the center of the cable, I 
pull upward, until the center of the cable is at the same height as 
the two ends of the cable. What upward force F am I applying at that 
instant?

Reference: AJP 58:1110 (1990). -Carl
-- 
Carl E Mungan, Assoc Prof of Physics  410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363
mailto:mungan@usna.edu    http://usna.edu/Users/physics/mungan/
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